The smallest value of $k$, for which both the roots of the equation $x^{2} - 8kx + 16(k^{2} - k + 1) = 0$ are real, distinct and have values at least $4$, is ________.

$\text{Given: } x^{2} - 8kx + 16(k^{2} - k + 1) = 0$

$\text{both roots real, distinct, and } \ge 4$

$\text{To find: the smallest } k \text{ satisfying all three conditions}$

$1.\ \text{find } D = b^{2}-4ac$

$2.\ \text{write the two roots}$

$3.\ \text{impose both roots } \ge 4$

$a=1,\quad b=-8k,\quad c=16(k^{2}-k+1)$

$\text{real and distinct} \Rightarrow D = b^{2}-4ac > 0$

$D = b^{2}-4ac = 64k^{2} - 64(k^{2}-k+1)$

$D = 64k^{2} - 64k^{2} + 64k - 64 = 64(k-1)$

$D>0 \Rightarrow 64(k-1)>0 \Rightarrow k>1$

$x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{8k \pm 8\sqrt{k-1}}{2}$

$x = 4k \pm 4\sqrt{k-1}$

$\text{smaller root} = 4k - 4\sqrt{k-1} \ge 4$

$\Rightarrow k - \sqrt{k-1} \ge 1$

$\text{larger root} = 4k + 4\sqrt{k-1} > \text{smaller root}$

$\text{let } t = \sqrt{k-1} \ge 0 \ \Rightarrow \ k = t^{2}+1$

$t^{2}+1 - t \ge 1 \Rightarrow t^{2}-t \ge 0$

$t(t-1) \ge 0$

$t \ge 0,\ t(t-1)\ge0 \Rightarrow t \ge 1$

$\sqrt{k-1} \ge 1 \Rightarrow k-1 \ge 1$

$\Rightarrow k \ge 2$

$k>1 \quad \text{and} \quad k \ge 2$

$\Rightarrow k \ge 2$

$k=2:\quad x^{2}-16x+48=0$

$x = \dfrac{16 \pm 8}{2} = 12,\ 4$

$\boxed{k_{\min} = 2}$

$\text{roots at } k=2:\ 4,\ 12 \ (\ge 4,\ \text{distinct})$